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Does a triangle with three right angles exist? Sure, on a sphere you just draw it around an octant. But can you make one on a flat sheet of paper?

  • The paper can be cut, folded or wrapped in any way provided that it remains smooth along the sides of the triangle and their immediate vicinity.

  • Stretching the paper to fit it around a ball is not allowed as this tears the paper and would violate the above rule. Similarly, shrinking it to fit a saddle would wrinkle it and thus not leave it smooth.

  • Any deformation you apply must assure that the vertex angles are in fact right angles. If you construct right angles initially, observing the first rule will guarantee this.

Hint:

Practically everyone loves ice cream wrapped in a thin edible waffle.

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  • $\begingroup$ The question seems to be inspired by this answer (spoiler warning). $\endgroup$ Commented Feb 2 at 4:31
  • $\begingroup$ Yes it is. But it turns out that answer is noylt unique for this question! $\endgroup$ Commented Feb 2 at 5:47

4 Answers 4

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The trick is to overlap two corners at a 90-degree angle, like a waffle cone:
enter image description here
Then you can draw a "triangle" across the overlap. Note that the red side will become a single straight line when the paper is curved.
enter image description here
enter image description here

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    $\begingroup$ A completely different solution from mine! See here for a solution that forms the convex triangle I envsioned. I did not require convex, so +1. $\endgroup$ Commented Feb 1 at 19:55
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My own solution is quite different from the first one posted, a non-uniqueness surprising to me!

In my solution we start with

a square drawn concentrically in a circular disk.

We

cut a 90° sector from the disk, shown below as including one full side of the square but in fact any 90°sector will do, if the vertex of the angle is exactly at the center.

Then simply

glue the cut edges together. The endpoints of the cut square will form the necessary triangle. In the picture below the angles are not depicted accurately due to limited drawing options on my device, but mathematically they are 90°. This triangle is also convex and equilateral.

Circle with smaller square in the center. A larger square turned 90 degrees has its left corner in the center of the circle and small square. An arrow is drawn to the right, pointing to a circle with a rounded-edges triangle inside.

The original idea behind this puzzle was a demonstration that although a conical surface looks Euclidean with triangles seeming to have angles that add up to 180°, that's not true if the triangle surrounds the apex. The singular point at the apex affects the space around it even though the space locally looks flat.

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On a flat piece of paper with a $135^\circ$ corner, we have the following triangle with right angled vertices ABC:

enter image description here

Note the part of the triangle on the top side of the paper is indicated in red, and the bottom in purple.

This is intended as a lateral thinking/humorous answer as although the flat paper is embedded smoothly in space, the triangle is actually drawn on the two sides of the paper, which is not.

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    $\begingroup$ The smoothness is dubious at the sharp edges. Make it like the surface of a beanbag, with the sharp boundaries replaced by cylindrical surfaces, and it will look more convincing. $\endgroup$ Commented Feb 2 at 11:47
  • $\begingroup$ @OscarLanzi You are right of course. I have edited to clarify my intention. As you say the answer could be modified to a mathematically correct one, but it would just be a messier version of the existing answers (+1)'s. $\endgroup$ Commented Feb 2 at 12:12
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Draw your three right angles - one on the pole and two on the hemisphere, of a sphere, forming a triangle. Now place your plane of paper through the sphere so it meets all three angles. The projection of your 3-right angled triangle on the paper is an equilateral triangle.

Now cut out the entire interior of your right angle triangle, and make two a creases in v-shapes from each corner to the edge of the paper.

It is now possible to move the three points where the paper meets the sphere in uniformly towards the centre, keeping them in contact with the sphere, all the while closing the creases and decreasing the angle between the paper and the sphere until they are tangential. Your paper now has a triangle with three right angles.

Note that they are right angles in 3d space, but they are not marked as right angles on the initial plane of paper, which may not have been the intention of the question, but I don't believe is explicitly ruled out by it.

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  • $\begingroup$ Your folds go right to the vertices and so miss the smoothness requirements all around the edges. $\endgroup$ Commented Feb 4 at 10:12
  • $\begingroup$ Thanks @OscarLanzi I'm guessing the meaning of smoothness to you is a precise mathematical topological meaning. This is possibly not fatal to my solution since the folds I described can be replaced with conical curves of the paper, which come to points at the corners of the triangle. Do these cones comply with your smoothness condition? $\endgroup$ Commented Feb 4 at 12:05
  • $\begingroup$ The conicap curves would work They would have to meet at folds in the center much like my beanbag description of another answer. What are the angles measured within the triangle (as distinct from through space? $\endgroup$ Commented Feb 4 at 13:03
  • $\begingroup$ @OscarLanzi it's an equilateral triangle on the paper (i.e. 60 deg) if you flatten it back out to a plane (assuming you draw it with 3-fold symmetry, which I only did for simplification by symmetry rather than due to seeing any lack of generality). $\endgroup$ Commented Feb 4 at 13:12
  • $\begingroup$ @OscarLanzi ...however your reply does suggest my solution (in a sense) might fail. If the curves would need to meet at folds in the very centre (as opposed to the interior) then this suggests the triangle would have to be contracted down to size zero for this to work, which may contradict your notion of a triangle. $\endgroup$ Commented Feb 4 at 13:15

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